There are 6 numbers x, y, 2, 5, 9, and 10. What is the range of the numbers?
1) The average (arithmetic mean) of x and y is 5
2) 2<x<y<10
==> In the original condition, there are 2 variables (x, y), and in order to match the number of variables to the number of equations, there must be 2 equations, and therefore C is most likely to be the answer. By solving con 1) and con 2), you get x+y=10, which becomes (x,y)=(3,7),(4,6)… Then, it is always range=10-2=8, so it is unique and sufficient. However, this statistics question is one of the key questions. So if you apply mistake type 4(A), for con 1), from x+y=10 and (x,y)=(3,7),(1,9), you always get range=8,9, and hence it is unique and sufficient. For con 2), from 2<x<y<10, you always get range=10-2=8, and hence it is unique and sufficient.
Therefore, the answer is B. This is a question related to mistake type 4(A).
Answer: B
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